Given that f (x) = - ax to the third power + 3 / 2x to the square - 2x, x = 1 is its extreme point, how to find the tangent equation of F (x) image at x = 3? (2) Find the maximum and minimum of F (x) in the interval [0,4].

Given that f (x) = - ax to the third power + 3 / 2x to the square - 2x, x = 1 is its extreme point, how to find the tangent equation of F (x) image at x = 3? (2) Find the maximum and minimum of F (x) in the interval [0,4].

When x = 1, the equation is equal to zero, and a = 2 / 3 can be obtained. The derivative equation has also been obtained, and the slope can be obtained by bringing 3 in

It is known that the derivative function f '(x) of function f (x) is a quadratic function, and the two of F' (x) = 0 are ± 1. If the sum of the maximum and minimum of F (x) is 0, f (- 2) = 2
(1) Finding the analytic expression of function f (x)
(2) If the function has maximum and minimum values in the open interval (m-9,9-m), find the value range of real number M
(3) Let f (x) = x * g (x), positive real numbers a, B, C satisfy Ag (b) = BG (c) = CG (a) > 0, and prove that a = b = C

(1) F '(x) is a quadratic function, and the two of F' (x) = 0 are ± 1. It can be seen that the integral of F '(x) = a (x + 1) (x-1) = ax ^ 2-A can be solved. The original function is f (x) = a (1 / 3 x ^ 3 - x) + C. as we know, x = 1, x = - 1 has extremum, and the sum of extremum is 0, then f (1) = - 2 / 3A + C, f (- 1) = 2 / 3A + C, f (- 1) + F (1) = 2C = 0

How to find the extreme point of quadratic function, y = ax & # 178; + BX + C

Hello
First, the function y = ax & # 178; + BX + C. is formulated
The form of y = a (x + m) ² + n
When a > 0, there is a minimum value in the function image
When a ＜ 0, the function image has a maximum value
The coordinates of the extreme point are (- m, n)
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Let the minimum value of function f (x) = x2-2x-1 in the interval [T, t + 1] be g (T), and find the range of G (T)

∵ f (x) = x2-2x-1 = (x-1) 2-2, ∵ axis of symmetry x = 1, vertex coordinates (1, - 2), as shown in the figure; f (x) monotonically decreases on (- ∞, 1), monotonically increases on (1, + ∞); when 0 ≤ t ≤ 1, G (T) = - 2; when t ≥ 1, G (T) = f (T) = t2-2t-1; when t ≤ 0, G (T) = f (T + 1) ）=(T + 1) 2-2 (T + 1) - 1 = t2-2.. g (T) = − 2 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 0 ≤ t ≤ 1t2 − 2T − 1 & nbsp; & nbsp; & nbsp; & nbsp; t > 1t2 − 2 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; t ＜ 0, the range of G (T) is [- 2, + ∞)